Re: Tangent of a point

Oh, dear. is a parabola, NOT any line at all! What you mean is that the derivative of so gives the slope of the tangent line.

Given that, at x= a, the derivative is and the value is . That means that the equation of the tangent line, at x= a, is . Saying that line goes through (0, -8) mean that x= 0, y= -8 makes that a true equation: solve .

Re: Tangent of a point

Originally Posted by minneola24

I attached the question number. I found the answer to A and B but I'm stuck on C.

For B I got y=x+7

but the tangent line for equation is 3x^2 -2x -4

Thank you.

for c,
remember that and lie on the same line,i.e the tangent line. So the slope of the line is .
Again you know that the slope of the (which you got by differentiating ) and thus for the slope is .
equate the slopes: and from the original equation:
so hence and